(navigation image)
Home American Libraries | Canadian Libraries | Universal Library | Community Texts | Project Gutenberg | Children's Library | Biodiversity Heritage Library | Additional Collections
Search: Advanced Search
Anonymous User (login or join us)
Upload
See other formats

Full text of "History Of The Theory Of Numbers - I"

CHAP, ill]                SYMMETRIC FUNCTIONS MODULO p.                         95
three primes, as well as P = 13 -37-73-457, for each of which the congruence holds for every a prime to P.
Welsch236 stated that if fc=4n+l is composite and <1000, 2*^=1 (mod k) only for A; = 561 and 645; hence nn=l (mod ft) for these two /b's.
P. Bachmann237 proved that x™~~l^l (mod pq) is never satisfied by all integers prime to pq if p and q are distinct odd primes [Carmichael235].
STMMETKIC FUNCTIONS OF 1, 2,. . .p — 1 MODULO p.
Report has been made above of the work on this topic by Lagrange,18 Lionnet,61 Tchebychef,75 Sylvester,78 Ottinger,92 Lucas,103 Cahen,125 Aubry,137 Ar^valo,150 Schuh,152 Frattini,153 Steiner,200 Jacobi,201 Hensel.204
We shall denote ln+2n+ . . . +(p — l)n by sn, and take p to be a prime.
E. Waring260 wrote a, /?, . . .for 1, 2, . . ., x, and considered
s = aapbvc . . . +ab/3 V . . . +aa/3 V ....
Ift-a+b+c+. . . is odd and <x, and x+l is prime, sis divisible by (x+l)2. Iit<2x and a, 6, . . .are all even and prime to 2s +1, s is divisible by 2x+l.
V. Bouniakowsky251 noted that sm is divisible by p2, if p> 2 and m is odd and not =1 (mod p — 1); also if both m==l (mod p — 1) and m^Q (mod p).
C. von Staudt252 proved that, if Sn(x) = l+2n+ . . . +zn,
Sn(ab) =bSn(a) +na/Sn_1(a)S1(&- 1)    (mod a2),
(mod a2).
•• integer.
If a, 6, . . . , I are relatively prime in pairs, Sn(ab...l)    SM
db.. .1           a                     I
A. Cauchy253 proved that 1 + 1/24-... +l/(p-l)=0 (mod p). G. Eisenstein254 noted that sm= — 1 or 0 (mod p) according as m is or is not divisible by p — 1.   If m, n are positive integers <p — 1,
2V"(cr+ir^O or - (p^m)    (mod p),
according as m+n<or^p —1.
L. Poinsot255 noted that, when a takes the values 1,..., p — 1, then (ax)n has the same residues modulo p as an, order apart. By addition, snzn=sn (mod p). Take x to be one of the numbers not a root of #n=l. Hence sn=0 (mod p) if n is not divisible by p — 1.
JML'interm&iiaire des math., 20, 1913, 94.
»"Archiv Math. Phys., (3), 21,1913, 185-7'.
MOMeditationes algebraicae, ed. 3, 1782, 382.
MlBull. Ac. Sc. St. P6terabourg, 4, 1838, 65-9.
MaJour. ftir Math., 21, 1840, 372-4.
'"M&n. Ac. Sc. de 1'Institut de France, 17, 1840, 340-1, footnote; Oeuvres, (1), 3, 81-2.
""Jour, fiir Math., 27, 1844, 292-3; 28, 1844, 232.
"•Jour, de Math., 10, 1845, 33-4.